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Raver32 writes "A team of astronomers announced they have discovered the smallest and potentially most Earth-like extrasolar planet yet. Five times as massive as Earth, it orbits a relatively cool star at a distance that would provide earthly temperatures as well, signaling the possibility of liquid water. 'The separation between the planet and its star is just right for having liquid water at its surface,' says astronomer and team spokesperson Stephane Udry of the Observatory of Geneva in Versoix, Switzerland. 'That's why we are a bit excited.' But researchers do not yet know if the planet contains water, if it is truly rocky like Earth, which might make it hospitable to life as we know it, or whether it is blanketed by a thick atmosphere. 'What we have,' Udry says, 'is the minimum mass of the planet and its separation" from its star.'"

I'm pretty sure that there is at least a triple dupe languishing somewhere in the Slashdot archives. But it's not raining hard enough for me to want to waste the time looking for it. Tell Guinness to hold that thought until next week.

You must be new here, this one is barely in the top ten list. I particularly like it when they manage to put a dupe on the front page at the same time as the original and then again a few days later...

I was just getting ready to post asking how often we are going to see teams announcing "the most Earth-like planet yet," but hopefully that won't happen after all. (It still may; science has gotten to love the press-release a bit too much, I think.)

Since Drake's equation needs to know the proportions of stars with planets, it would require us to have known negative results as well as known positives in order for it to give any meaningful results.

At the moment, we can say there are a few hundred planets, out of maybe a few thousand stars that we've scanned, but for the stars where we haven't found anything, we don't know for sure whether that's because there isn't anything there, or because we just aren't looking hard enough.

Forget the proportion of stars with planets. Fl is the real unknown. Why assume all planets that can support life will develop life? What if life is actually pretty rare? Try plugging in values less that 1 for Fl (0.1, 0.01, 0.001) and you'll get some disappointing results from this equation. Trying to quantify something with so many unknowns seems pretty silly to me. On the other hand maybe life isn't rare, but that's just me being hopeful.

From the blurb itself, it's five time the size of earth, it's revolving around a cooler sun than earth, and it might not have liquid water or a thick atmosphere. Yeah, that's exactly like earth!

You're missing the point. By Earth-like they mean telluric planet, as in, not a gas giant. That's all. And that matters because until now we haven't found so many of them, most of the planets we've found were gas giants orbiting close to their star. But as time goes by we find ever decreasingly large planets that get closer and closer to the Earth in size.

It is located roughly at a point where it COULD have liquid water and it could have a thick atmosphere.It may or may not.But may is better than not a chance.Seems like it would be worth pointing a radio telescope at and see if we find there version of I Love Lucie.

Assuming that the density is a little bit less than the Earth (more like the Moon or Mars) and this "Super Earth" is thus larger by a sizable fraction..... what is the geological environment of a planet such as this like?

Since the interior heat of this planet has less surface area in proportion to its volume, internal heat from its formation and nuclear decay from heavy elements (like Uranium) would therefore cause a much larger interior heat sink... and causing substantially more techtonic activity and a great many more volcanoes.

Using Mars as a comparator here as well, Mars is smaller than the Earth, and geologically dead, with fewer but much larger volcanoes. Is it reasonable to assume this planet... if it had a rocky "surface", would literally be littered with smaller volcanoes over nearly all of its surface with much smaller "continents"?

Assume that the age of this planet is roughly similar to that of the Earth and that heavy metals (heavier than Iron) in its formation are roughly proportional to what we find on the Earth.

I just don't find that this would be all that pleasant of a place to be at, and the nearly constant volcanism would IMHO kill off nearly any attempt to colonize this planet with life.

The latest results of Messenger's first flyby of Mercury confirms a magnetic field and molten outer core. Conversely, Venus which is Earth's twin in size, seems a lot more dead. A more important factor may have been chemical composition at the time of formation - Mercury had more metal. Elements may have been unevenly distributed as function of distance from the Sun in the original planetary nebula.

To the many, many people who've taken the time to correct my shitty assumption, berate me, mod me down, and otherwise point out that in my rush to post I forgot to turn my brain on... I hang my head in shame. I will now seek out my grade 10 physics teacher (or locate his grave as the case may be) and confess my sins.

And of course, for the angriest among you, this post presents another opportunity to mod me down.

If we took all the stuff Earth is made of, took it apart and measured the average density of all those rocks at 1 atm, we would get a significantly lower average than what we get by dividing the estimated mass of Earth by its estimated volume.

Hm, Slashdot ate my reply which said pretty much what you just said, except that I added that this means that larger planets are almost always denser in spite of the fact that the terrestrial planets have nearly the same composition. (Shut up, Mercury.) A planet larger than Earth should be somewhat denser, so the assumption I made above isn't exactly true. It is, however, probably good enough.

Not necessarily. You'll also get a denser atmosphere, which should alter the dynamics of flying a bit. Our birds might not be able to make it, but I wouldn't be surprised if bird-analogues couldn't evolve there.

When they say "nearly the size of earth" they're speaking in an astronomical scale, which would qualify something 5-times as large as the earth as 'nearly'. It's not composed of gold or other heavy metal.

It's a poorly written and shite article, but the box off to the side says:

One of two newly discovered exoplanets is nearly the size of Earth...

So, assuming they're talking about the same one, it should be roughly 5 times our gravity.

Not so. If the planet has twice the diameter of earth, that falls well within the category of "nearly the size of Earth" for astronomers. Since gravity decreases proportionally to the square of the distance, gravity would be only 5/(2^2) times as strong as on Earth, an increase of a mere 20%.

If it has approximately the same density as earth, then since volume of a sphere increases proportionally to the cube of the radius/diameter, it would have 5^(1/3) times as large a diameter as earth, which is about 1.71 -- even closer to the size of our Earth. It would also wind up with gravity 1.71 times as strong, since 5/((5^(1/3))^2) == 5/(5^(2/3)) == 5^(1/3).

you're both missing the point, if the planet has water, then the only factor is compression ratio, aquatic life don't suffer from gravity like land bound creatures do, if they have neutral buoyancy the only effect of gravity they feel is the relative pressure of the water at the depth they live in.

considering there are whales that can go very deep in the ocean, to the very surface, the pressure regulation seems to be easily solved.

gravity only becomes an issue when life tries to evolve from aquatic life to

I think that in planetary terms we can safely assume 5x mass will create an environment of roughly 5g... maybe give or take 20%. Enough to ensure that the simple act of getting out of bed would be a gruelling ordeal.

Another problem I noticed after actually reading TFA:

Gliese 581 c, orbits at one fourteenth the distance between Earth and the sun. But the red dwarf is 50 times cooler than the sun. The group estimates that the planet would experience temperatures in the zero-to-40-degree-Celsius (32-to-104-Fahrenheit) range.

It is my understanding that red dwarfs, while generating reduced heat and light output, produce solar flares that are almost as intense as those produced by a G class star. So if a planet exists in the habitable zone it is also exposed to periodic sterilizing blasts of charged particles.

Maybe if we're lucky the planet happens to have a really strong magnetic field... then we just have the crushing g load to contend with.

"Space is big. Really big. You just won't believe how vastly, hugely, mind-bogglingly big it is."...and thats the thing about a really big place- sheer size provides an astronomical number of opportunities for such astronomically unlikely things to happen, over and over agin.

I think that in planetary terms we can safely assume 5x mass will create an environment of roughly 5g... maybe give or take 20%. Enough to ensure that the simple act of getting out of bed would be a gruelling ordeal.

I don't know about you, but getting out of bed is a grueling ordeal at any gravity for me!

I think that in planetary terms we can safely assume 5x mass will create an environment of roughly 5g... maybe give or take 20%. Enough to ensure that the simple act of getting out of bed would be a gruelling ordeal.

Another problem I noticed after actually reading TFA:

No, it would not. It would need to be much denser than Earth for that to happen. This is basically impossible for an object of that mass.

Assuming roughly Earth like density (which is quite plausible), Radius will scale like Mass to the 1/3, while gravity scales like mass / radius squared. This works out to about 1.7 times Earth gravity at the surface.

Let's assume that the average density of the earth-like planet is the same as Earth. (It wouldn't be an earth like planet if it were significantly different.) Then we can use the volume of the sphere to relate the mass and the surface radius. Since M = 4/3 * \pi * R^3 * \rho, where \rho is the density, it is easy to see that the surface radius goes like the cube root of the mass. Putting this into Newton's equation, we can see that a = GM/R^2 means that the surface gravity is also going to go like the cube root of the mass. If the mass is five times that of Earth, then the surface gravity will be the cube root of 5 greater than Earth's or about 1.7 times Earth normal.

Taking differences in the mean density into account is no more difficult, but I leave that as an exercise for the reader.

Even without doing any maths, some basic sanity checking can tell the grandparent that he's an idiot. Gravity on the surface of Jupiter is about 2.3 times that on Earth and Jupiter is a shade under 320 times the mass of the Earth. A planet with five times the surface gravity of Earth would have to be incredibly dense.

To be pedantic, Jupiter doesn't have a surface. It just has a really thick atmosphere that turns into molten stuff when you get deep enough. It's not a stretch to say there's nothing solid in the entire planet.

I think that in planetary terms we can safely assume 5x mass will create an environment of roughly 5g... maybe give or take 20%.

How do you justify that remark? Mars [wikipedia.org] has a mass 1/9 of Earth's but a surface gravity over 1/3 of Earth's. Mercury [wikipedia.org] has a mass 1/18 that of Earth but has gravity slightly higher than that of Mars.

There's just no way you can have confidence within 20% that the gravity will be proportional to the mass.

The problem is the relative brightness of the parent star. Even if (as in this case) it's a red dwarf, it's still about a bazillion times brighter than the reflected light reaching earth from the planet itself.

As others have said, if the planet has the same average density as the earth, then its surface gravity would be 1.7 times greater. It's interesting, however, to see what how the density/surface gravity depends on composition.

In this paper [arxiv.org] there are theoretical relations between planet radius and mass for a wide range of possible planet compositions. These are computed using equations of state that are largely determined from laboratory experiments.

You're probably thinking of the shell theorem, which says that a uniform sphere of mass is gravitationally equivalent to a point mass located at the center of the sphere. This theorem does imply that a larger radius = less gravity at the surface.

To calculate the gravitational effect of a massive sphere, its whole mass can be
considered accumulated in its center as long as you are outside of it.
So the gravitational acceleration indeed only depends on mass an distance.
Mathematical fact.

Neat additional trivia:
- Inside a hollow sphere, there is no gravitational effect by the sphere's mass - it cancels out exactly.
That's why
- Inside a massive sphere, gravitational acceleration increses linearly with the radial distance to the center.
(the mass increases with r^3 as you get further out, its effect decreases by 1/r^2 - and as it can be considered
concentrated in the middle, you get an increase by a factor of r^3/r^2 = r
Gravity is fun:-)

``No, we can even measure how surface gravity varies from g=9.78 m/s2 to g=9.82 m/s2 when moving from the equator towards a pole. And this is because Earth is not perfectly round, the people at higher latitudes are closer to the center of Earth and fall faster.''

Actually, if I remember correctly, an object weighs less at the equator because the earth is rotating. An object in motion will travel in a straight line unless acted on by an outside force. Therefore, at the equator, gravity also acts as a centripetal force to keep the object from flying off into space. If the object is near the poles, less centripetal force is required to keep it from flying away and it is held tighter to the surface.

It may be earthlike, but it sure wouldn't be a comfortable place to spend any amount of time.

Mass alone says very little about the surface gravity of a planet - you need to know the radius of the object to make any statement about its surface gravity. Earth's moon has slightly over a percent of the mass of Earth, but about 1/6g surface gravity. Mars has only about 10% of the mass of Earth, while having 1/3g surface gravity.

Acceleration due to gravity scales as mass over Radius squared, whereas Radius will scale as Mass to the one third, assuming relatively constant density. So, the radius would be about 1.7 times that of Earth.

Thus, acceleration due to gravity would be about 5 / (1.7 ^ 2) or about 1.7 times Earth acceleration (10 m / s ^ 2). This is all assuming it is Earth like in composition, which we don't know for sure.

1.7 times Earth gravity would be pretty high, but it might be livable. It is worth noting, howe

I think the real problem to make it earth-like is not surface gravity, but instead the distance at which it can still trap its atmosphere. The much thicker atmosphere probably means that the planet traps heat from its "sun" very efficiently.

If there is complex land life I sure don't want to meet it. as for it could survive it would need to be super strong, and light. So an humanoid would be like 25 lbs and probably on the average twice as strong as us.

If we assume this planet is truly "earth-like" and has similar density, then it is not a true statement that it will have 5 times the gravity of earth. If the density is similar to that of Earth, then the size of the planet will be larger. The radius will be larger by a factor of the cube root of 5 (the real one, that is), which is about 1.7, which is also almost exactly the square root of 3.

That's unlikely, and as a couple other responders have pointed out, it depends on the radius of the planet. If it has the same average density as the Earth, then the gravitational forces at its surface are 5^(1/3)=1.7 times as strong.

Surface gravity is proportional to the cube root of the mass if you assume similar composition, because it's proportional to the planet's mass and inversely proportional to the distance from the center of mass to the surface (i.e. the planet's radius). The radius will be 5^(1/3) = 1.71x. The gravity will then be 5/(1.71^2) = 1.71x.

You know...I know your joking, but even if we were to send someone over there, at 20 light years away, the amount of time it would take to receive the transmitted data would take a life time to get back. (Do communication waves travel the speed of light?)

You mean dead, desiccated prisoners. How long would it take Voyager to travel 20 ly? And we want to send people on that trip? Do we tell them we're going to use a worm hole/star gate/warp drive on the ship and that's why they're only getting a single soda and pack of peanuts for supplies?

There are two press releases that I really really wish I could take a vacation from.
First, announcement of an "earth-like planet" that, when you read the details, isn't actually earthlike.
Second, announcement of a "breakthrough in solar cell technology" that, when you read the details, is a piece of fundamental research that may be interesting in a scientific way but the researchers have never actually produced a working solar cell.

indeed... see Mars... it's internal warmth died off quickly, being not only a small, but light-for-its-size, planet... Earth is waning now, but remained internally hot for billions of years longer... and even in spite of that, it once spent a fair period of time as an icecube before the cambrian explosion... just goes to show that location is important, but not enough.